The warranty on a machine specifies that it will be replaced at failure or age 4, whichever occurs first. the machines age at failure, x, has density function:

f(x) = 1/5 for 0 0 otherwise

let y be the age of the machine at the time of replacement.
determine the variance of y.

Step-by-step explanation:

Previous concepts

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).

Solution to the problem

Since the warranty on a machine specifies that it will be replaced at failure or age 4 and the distribution for X is defined between 0 and 5 then if we define the random variable Y ="the age of the machine at the time of replacement" we know that the values for Y needs to be between 0 and 4 or between 4 and and we can define the following density function:

for other case

Now we can apply the definition of expected value and we have this:

And for the second moment we have:

And the variance would be given by:

this linear equation would have just one solution.

step-by-step explanation:

the reason why this would have one solution is that it is an expression with just one variable. since that variable is not raised to a power, it will always have just one solution.